\(2013=2.1000+10+3=2.10^3+1.10^1+3.10^0\)

\(20112013=2.10000000+100000+10000+2.1000+10+3\)

\(=2.10^7+10^5+10^4+2.10^3+10^1+3.10^0\)

2013 = 2 x 1000 + 10 + 3 = 2 x 10^3 + 1 . 10^1 + 3 . 10^0

20112013 = 2 x 10000000 + 100000 + 2 x 1000 + 10 + 3 =  2 x 10^7 + 10^5 + 10^4 + 2 x 10^3 + 10^1 + 3 x 10^0

ok nhé với lại 10^2 là 10 mũ 2 nhé =>

a) \(8=2^3\)

\(16=4^2\)

\(27=3^3\)

\(81=9^2\)

\(100=10^2\)

b) \(1000=10^3\)

\(1,000,000=10^6\)

\(1,000,000,000=10^9\)

100.000 } 12 chữ số 0 = 10^12

9¹² x 27⁵ = (3²)¹² x (3³)⁵ = 3²⁴ x 3¹⁵ = 3²⁴⁺¹⁵ = 3³⁹

Chọn D

D. 3³⁹

Ta có:

 \(9^{12}.27^5=\left(3^2\right)^{12}.\left(3^3\right)^5=3^{2.12}.3^{3.5}=3^{24}.3^{15}=3^{24+15}=3^{39}\)

Vậy đáp án là D

Bài 6: 

a: \(2^{27}=8^9\)

\(3^{18}=9^9\)

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)

Ta có: 712 : 74 = 7(12-4) = 78

a: \(3\cdot3\cdot3\cdot3\cdot3=3^5\)

b: \(y\cdot y\cdot y\cdot y=y^4\)

c: \(5\cdot p\cdot5\cdot p\cdot2\cdot q\cdot4\cdot q=25\cdot2\cdot4\cdot p^2q^2=2\cdot\left(10qp\right)^2\)

d: \(a\cdot a+b\cdot b+c\cdot c+d\cdot d\cdot d\cdot d=a^2+b^2+c^2+d^4\)

a) 3.3.3.3.3 = 3\(^5\)

b) y.y.y.y = y\(^4\)

c) 5.p.5.p.2.q.4.q = 5\(^2\).p\(^2\).q\(^2\).2\(^3\) 

(2\(^3\) ở đây là vì 2.4 = 2.2.2 = 2\(^3\))

d) a.a + b.b + c.c.c + d.d.d.d = a\(^2\)+b\(^2\)+c\(^3\)+d\(^4\)

1.
a) \(3^4\times3^5\times3^6=3^{4+5+6}=3^{15}\)

b) \(5^2\times5^4\times5^5\times25=5^2\times5^4\times5^5\times5^2=5^{2+4+5+2}=5^{13}\)

c) \(10^8\div10^3=10^{8-3}=10^5\)

d) \(a^7\div a^2=a^{7-2}=a^5\)

2.

\(987=900+80+7\\ =9\times100+8\times10+7\\ =9\times10^2+8\times10^1+7\times10^0\)

\(2021=2000+20+1\\ =2\times1000+2\times10+1\times1\\ =2\times10^3+2\times10^1+1\times10^0\)

\(abcde=a\times10000+b\times1000+c\times100+d\times10+e\times1\\ =a\times10^4+b\times10^3+c\times10^2+d\times10^1+e\times10^0\)

còn bài 2

Ta có: a4 : a4 = a(4-4) = a0

x6 : x3 = x(6-3) = x3

\(\dfrac{1}{32}\) = 2^-5